Question: Simplify and expand the following expression: $ \dfrac{1}{r - 6}+ \dfrac{2}{4r - 20}+ \dfrac{2r}{r^2 - 11r + 30} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the second term: $ \dfrac{2}{4r - 20} = \dfrac{2}{4(r - 5)}$ We can factor the quadratic in the third term: $ \dfrac{2r}{r^2 - 11r + 30} = \dfrac{2r}{(r - 6)(r - 5)}$ Now we have: $ \dfrac{1}{r - 6}+ \dfrac{2}{4(r - 5)}+ \dfrac{2r}{(r - 6)(r - 5)} $ The least common multiple of the denominators is: $ (r - 6)(r - 5)$ In order to get the first term over $(r - 6)(r - 5)$ , multiply by $\dfrac{4(r - 5)}{4(r - 5)}$ $ \dfrac{1}{r - 6} \times \dfrac{4(r - 5)}{4(r - 5)} = \dfrac{4(r - 5)}{(r - 6)(r - 5)} $ In order to get the second term over $(r - 6)(r - 5)$ , multiply by $\dfrac{r - 6}{r - 6}$ $ \dfrac{2}{4(r - 5)} \times \dfrac{r - 6}{r - 6} = \dfrac{2(r - 6)}{(r - 6)(r - 5)} $ In order to get the third term over $(r - 6)(r - 5)$ , multiply by $\dfrac{4}{4}$ $ \dfrac{2r}{(r - 6)(r - 5)} \times \dfrac{4}{4} = \dfrac{8r}{(r - 6)(r - 5)} $ Now we have: $ \dfrac{4(r - 5)}{(r - 6)(r - 5)} + \dfrac{2(r - 6)}{(r - 6)(r - 5)} + \dfrac{8r}{(r - 6)(r - 5)} $ $ = \dfrac{ 4(r - 5) + 2(r - 6) + 8r} {(r - 6)(r - 5)} $ Expand: $ = \dfrac{4r - 20 + 2r - 12 + 8r}{4r^2 - 44r + 120} $ $ = \dfrac{14r - 32}{4r^2 - 44r + 120}$ Simplify: $ = \dfrac{7r - 16}{2r^2 - 22r + 60}$